# Thread: Focus distances in intervals

1. ## Focus distances in intervals

Problem above, I use the distance formula from the focus to the intervals(0/20/40/60/80/100).
I use x2=4(1/4)y as the formula but end up with answers far too large using the distance formula.
Where am I going wrong?

finding a formula on my calculator(guessing and checking parabolas that fit into the given coords)
gives me the solutions, is there no other way to solve without a calculator?Feels like I'm cheating
formula is y=.004x^2

2. ## Re: Focus distances in intervals

Hello, Greymalkin!

The problem does not involve the focus.

The cables of a suspension bridge are 50 ft above the roadbed at the ends of the bridge,
and 10 ft above it in the center of the bridge. .The bridge is 200 ft long.
Vertical cables are to be spaced every 20 ft along the bridge.
Calculate the lengths of these vertical cables.

Code:
                |
*         |         *
:         |         :
:*        |        *:
: *       |       * :
:   *     |     *   :
50 :       * * *       : 50
:         |         :
:         |10       :
:         |         :
. - - * - - - - + - - - - * - -
-100        |        100
Answers are: 10, 11.6, 16.4, 24.4, 35.6, 50

I placed the parabola on a coordinate system.

The roadbed is the $x$-axis.
The tops of the bridge are at $(\pm100,50)$
The lowest point is at $(0,10)$

The equation has the form: . $y \:=\:ax^2 + b$

Substitute $(0,10)\!:\;10 \:=\:a(0^2) + b \quad\Rightarrow\quad b \,=\,10$

Substitute $(100,50)\!:\;50 \:=\:a(100^2) + 10 \Rightarrow\quad 10,000a \:=\:40 \quad\Rightarrow\quad a \:=\:\tfrac{1}{250}$

Hence, the parabola is: . $y \:=\:\tfrac{1}{250}x^2+ 10$

Now, evaluate $y$ for $x \:=\:0,\pm20,\pm40,\pm60, \pm80, \pm100.$

3. ## Re: Focus distances in intervals

Thanks again Soroban, believe it or not I feel much better knowing I only slightly cheated.